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course Phy 121
The qa 01 could not be opened.
ph1 query 1 #$&*
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Question: `qExplain in your own words how the standard deviation of a set of numbers is calculated.
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Your solution:
This is a long process to do without the use of a scientific calculator(which is the only way that I know for sure how to do it). I'll explain the way that I know using my calculator.
First, I go to the list operation. I enter the numbers in the vertical column. Next, I select STATISTICS from a drop down menu, then STAT CALCULATIONS. Finally, I choose ONE-VARIABLE STATISTICS and the calculator will display the MEAN, STANDARD DEVIATIONS, etc.
I know that the standard deviation is calculated by first finding the mean, then doing several different averages by dividing sums of numbers by the number of trials but I'm not confident that I have the experience to do it right now, without my calculator.
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That's a good explanation of the calculator steps. However it's not an explanation of the mathematics.
As I believe you realize, the calculator does not provide not a mathematical way to explain the process of finding a quantity. It's handy for calculating the quantity, but calculator steps are never an explanation of a mathematical operation.
The same goes for spreadsheets. Very handy, very useful, but not explanatory.
The process is part of one of the lab exercises in this section.
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confidence rating #$&*: 3
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Question: State the given definition of the average rate of change of A with respect to B.
Briefly state what you think velocity is and how you think it is an example of a rate of change.
In reference to the definition of average rate of change, if you were to apply that definition to get an average velocity, what would you use for the A quantity and what would you use for the B quantity?
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Your solution:
average rate = change in A / change in B
The average velocity of an object between two clock times is the average rate at which its position changes between those clock times. The average rate at which position changes is ave rate = change in position / change in clock time = ave velocity.
I would use change in position as the A quantity and I would use the clock time as the B quantity.
confidence rating #$&*: 3
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Given Solution:
A rate is a change in something divided by a change in something else.
This question concerns velocity, which is the rate of change of position: change in position divided by change in clock time. **
NOTE ON NOTATION
Students often quote a formula like v = d / t. It's best to avoid this formula completely.
The average velocity on an interval is defined as the average rate of change of position with respect to clock time. By the definition of average rate, then, the average velocity on the interval is v_ave = (change in position / change in clock time).
• One reason we might not want to use v = d / t: The symbol d doesn't look like a change in anything, nor does the symbol t. Also it's very to read 'd' and 'distance' rather than 'displacement'.
• Another reason: The symbol v doesn't distinguish between initial velocity, final velocity, average velocity, change in velocity and instantaneous velocity, all of which are important concepts that need to be associated with distinct symbols.
In this course we use `d to stand for the capital Greek symbol Delta, which universally indicates the change in a quantity. If we use d for distance, then the 'change in distance' would be denoted `dd. It's potentially confusing to have two different d's, with two different meanings, in the same expression.
We generally use s or x to stand for position, so `ds or `dx would stand for change in position. Change in clock time would be `dt. Thus
v_Ave = `ds / `dt
(or alternatively, if we use x for position, v_Ave = `dx / `dt).
With this notation we can tell that we are dividing change in position by change in clock time.
For University Physics students (calculus-based note):
If x is the position then velocity is dx/dt, the derivative of position with respect to clock time. This is the limiting value of the rate of change of position with respect to clock time. You need to think in these terms.
v stands for instantaneous velocity. v_Ave stands for the average velocity on an interval.
If you used d for position then you would have the formula v = dd / dt. The dd in the numerator doesn't make a lot of sense; one d indicates the infinitesimal change in the other d.
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Self-critique (if necessary):
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Question: Given average speed and time interval how do you find distance moved?
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Your solution:
change in position = speed * time Remembering that the speed is an average, not instantaneous. So if I am traveling at an average speed of 50 miles per hour for 2 hours then I've moved 100 miles.
confidence rating #$&*: 3
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Given Solution:
** You multiply average speed * time interval to find distance moved.
For example, 50 miles / hour * 3 hours = 150 miles. **
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Question: Given average speed and distance moved how do you find the corresponding time interval?
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Your solution:
I would manipulate the formula to solve for time. distance moved = avg speed * time interval,
therefore time interval = distance moved/avg speed
From the previous example, if I moved 100 miles at an average speed of 50 miles per hour, then it took me 2 hours.
confidence rating #$&*: 3
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Given Solution:
** time interval = distance / average speed. For example if we travel 100 miles at 50 mph it takes 2 hours--we divide the distance by the speed.
In symbols, if `ds = vAve * `dt then `dt = `ds/vAve.
Also note that (cm/s ) / s = cm/s^2, not sec, whereas cm / (cm/s) = cm * s / cm = s, as appropriate in a calculation of `dt. **
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Question: Given time interval and distance moved how do you get average speed?
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Your solution:
v_Ave = distance moved / time interval
If I drive 100 miles and it takes me 2 hours, then my average speed is 50 miles per hour.
confidence rating #$&*: 3
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Given Solution:
** Average speed = distance / change in clock time. This is the definition of average speed.
For example if we travel 300 miles in 5 hours we have been traveling at an average speed of 300 miles / 5 hours = 60 miles / hour. **
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Question: A ball rolls from rest down a book, off that book and onto another book, where it picks up speed before rolling off the end of that book. Consider the interval that begins when the ball first encounters the second book, and ends when it rolls of the end of the book.
For this interval, place in order the quantities initial velocity (which we denote v_0), and final velocity (which we denote v_f), average velocity (which we denote v_Ave).
During this interval, the ball's velocity changes. It is possible for the change in its velocity to exceed the three quantities you just listed? Is it possible for all three of these quantities to exceed the change in the ball's velocity? Explain.
Note that the change in the ball's velocity is denoted `dv.
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Your solution:
I'm having a little trouble understanding the question. I'll assume that what is meant by exceeding the three quantities listed means the values for v_0, v_f, and v_Ave. I would say yes because the change in velocity of an object can be constant, decreasing, or increasing at given points on a graph. So, the velocity could, at some point on the book, be higher than the either the initial, final, or average velocity's. It would help to know it the second book has any elevation or if it's laying flat. If it's elevated, then I would suggest that the balls final velocity would be the largest number as it would be constantly accelerating while in contact with the second book. Now that I have reasoned this way, I would suggest that while the velocity does change on the second book, it will never be higher at any given point in between the initial contact with the second book and the final point of contact with the book as it will be AT the final point of contact with the book. So my final answer is no. The largest value for change in velocity is at the v_f, the point at which is last contacts the surface of the book. It doesn't slow down at any point while in contact with the second book.
confidence rating #$&*: 3
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Question: If the velocity at the beginning of an interval is 4 m/s and at the end of the interval it is 10 m/s, then what is the average of these velocities, and what is the change in velocity?
List the four quantities initial velocity, final velocity, average of initial and final velocities, and change in velocity, in order from least to greatest.
Give an example of positive initial and final velocities for which the order of the four quantities would be different.
For positive initial and final velocities, is it possible for the change in velocity to exceed the other three quanities?
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Your solution:
The average is 4 + 10 = 14/2 = 7 m/s. The change in velocity is 10 - 4 = 6 m/s.
Initial velocity: 4 m/s
Change in velocity: 6 m/s
Ave velocity: 7 m/s
Final velocity: 10 m/s
An example would be to switch the situation: Initial velocity 10 m/s, final velocity 4 m/s. Now, the final velocity is first and the initial velocity is last.
confidence rating #$&*: 3
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Question: If the position of an object changes by 5.2 meters, with an uncertainty of +-4%, during a time interval of 1.3 seconds, with an uncertainty of +-2%, then
What is the uncertainty in the change in position in meters>
What is the uncertainty in the time interval in seconds?
What is the average velocity of the object, and what do you think ia the uncertainty in the average velocity?
(this last question is required of University Physics students only, but other are welcome to answer): What is the percent uncertainty in the average velocity of the object, and what is the uncertainty as given in units of velocity?
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Your solution:
The uncertainty is 0.208 meters
The uncertainty in the time interval is 0.026 seconds
The average velocity is 5.2m/1.3s=4m/s with an uncertainty of +-5.3%(or 0.212)
confidence rating #$&*: 3
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Self-critique (if necessary):
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Self-critique rating:
@&
What is the maximum possible distance?
What is the minimum possible time interval?
What do you get when you calculate velocity based on these quantities?
By what percent does this result differ from the 4 m/s velocity you calculated?
*@
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Question: A ball rolls from rest down a book, off that book and onto another book, where it picks up speed before rolling off the end of that book. Consider the interval that begins when the ball first encounters the second book, and ends when it rolls of the end of the book.
For this interval, place in order the quantities initial velocity (which we denote v_0), and final velocity (which we denote v_f), average velocity (which we denote v_Ave).
During this interval, the ball's velocity changes. It is possible for the change in its velocity to exceed the three quantities you just listed? Is it possible for all three of these quantities to exceed the change in the ball's velocity? Explain.
Note that the change in the ball's velocity is denoted `dv.
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Your solution:
I'm having a little trouble understanding the question. I'll assume that what is meant by exceeding the three quantities listed means the values for v_0, v_f, and v_Ave. I would say yes because the change in velocity of an object can be constant, decreasing, or increasing at given points on a graph. So, the velocity could, at some point on the book, be higher than the either the initial, final, or average velocity's. It would help to know it the second book has any elevation or if it's laying flat. If it's elevated, then I would suggest that the balls final velocity would be the largest number as it would be constantly accelerating while in contact with the second book. Now that I have reasoned this way, I would suggest that while the velocity does change on the second book, it will never be higher at any given point in between the initial contact with the second book and the final point of contact with the book as it will be AT the final point of contact with the book. So my final answer is no. The largest value for change in velocity is at the v_f, the point at which is last contacts the surface of the book. It doesn't slow down at any point while in contact with the second book.
confidence rating #$&*: 3
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Question: If the velocity at the beginning of an interval is 4 m/s and at the end of the interval it is 10 m/s, then what is the average of these velocities, and what is the change in velocity?
List the four quantities initial velocity, final velocity, average of initial and final velocities, and change in velocity, in order from least to greatest.
Give an example of positive initial and final velocities for which the order of the four quantities would be different.
For positive initial and final velocities, is it possible for the change in velocity to exceed the other three quanities?
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Your solution:
The average is 4 + 10 = 14/2 = 7 m/s. The change in velocity is 10 - 4 = 6 m/s.
Initial velocity: 4 m/s
Change in velocity: 6 m/s
Ave velocity: 7 m/s
Final velocity: 10 m/s
An example would be to switch the situation: Initial velocity 10 m/s, final velocity 4 m/s. Now, the final velocity is first and the initial velocity is last.
confidence rating #$&*: 3
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Question: If the position of an object changes by 5.2 meters, with an uncertainty of +-4%, during a time interval of 1.3 seconds, with an uncertainty of +-2%, then
What is the uncertainty in the change in position in meters>
What is the uncertainty in the time interval in seconds?
What is the average velocity of the object, and what do you think ia the uncertainty in the average velocity?
(this last question is required of University Physics students only, but other are welcome to answer): What is the percent uncertainty in the average velocity of the object, and what is the uncertainty as given in units of velocity?
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Your solution:
The uncertainty is 0.208 meters
The uncertainty in the time interval is 0.026 seconds
The average velocity is 5.2m/1.3s=4m/s with an uncertainty of +-5.3%(or 0.212)
confidence rating #$&*: 3
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
#$&*
"
Self-critique (if necessary):
------------------------------------------------
Self-critique rating:
@&
What is the maximum possible distance?
What is the minimum possible time interval?
What do you get when you calculate velocity based on these quantities?
By what percent does this result differ from the 4 m/s velocity you calculated?
*@
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Question: A ball rolls from rest down a book, off that book and onto another book, where it picks up speed before rolling off the end of that book. Consider the interval that begins when the ball first encounters the second book, and ends when it rolls of the end of the book.
For this interval, place in order the quantities initial velocity (which we denote v_0), and final velocity (which we denote v_f), average velocity (which we denote v_Ave).
During this interval, the ball's velocity changes. It is possible for the change in its velocity to exceed the three quantities you just listed? Is it possible for all three of these quantities to exceed the change in the ball's velocity? Explain.
Note that the change in the ball's velocity is denoted `dv.
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Your solution:
I'm having a little trouble understanding the question. I'll assume that what is meant by exceeding the three quantities listed means the values for v_0, v_f, and v_Ave. I would say yes because the change in velocity of an object can be constant, decreasing, or increasing at given points on a graph. So, the velocity could, at some point on the book, be higher than the either the initial, final, or average velocity's. It would help to know it the second book has any elevation or if it's laying flat. If it's elevated, then I would suggest that the balls final velocity would be the largest number as it would be constantly accelerating while in contact with the second book. Now that I have reasoned this way, I would suggest that while the velocity does change on the second book, it will never be higher at any given point in between the initial contact with the second book and the final point of contact with the book as it will be AT the final point of contact with the book. So my final answer is no. The largest value for change in velocity is at the v_f, the point at which is last contacts the surface of the book. It doesn't slow down at any point while in contact with the second book.
confidence rating #$&*: 3
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Question: If the velocity at the beginning of an interval is 4 m/s and at the end of the interval it is 10 m/s, then what is the average of these velocities, and what is the change in velocity?
List the four quantities initial velocity, final velocity, average of initial and final velocities, and change in velocity, in order from least to greatest.
Give an example of positive initial and final velocities for which the order of the four quantities would be different.
For positive initial and final velocities, is it possible for the change in velocity to exceed the other three quanities?
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Your solution:
The average is 4 + 10 = 14/2 = 7 m/s. The change in velocity is 10 - 4 = 6 m/s.
Initial velocity: 4 m/s
Change in velocity: 6 m/s
Ave velocity: 7 m/s
Final velocity: 10 m/s
An example would be to switch the situation: Initial velocity 10 m/s, final velocity 4 m/s. Now, the final velocity is first and the initial velocity is last.
confidence rating #$&*: 3
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Question: If the position of an object changes by 5.2 meters, with an uncertainty of +-4%, during a time interval of 1.3 seconds, with an uncertainty of +-2%, then
What is the uncertainty in the change in position in meters>
What is the uncertainty in the time interval in seconds?
What is the average velocity of the object, and what do you think ia the uncertainty in the average velocity?
(this last question is required of University Physics students only, but other are welcome to answer): What is the percent uncertainty in the average velocity of the object, and what is the uncertainty as given in units of velocity?
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Your solution:
The uncertainty is 0.208 meters
The uncertainty in the time interval is 0.026 seconds
The average velocity is 5.2m/1.3s=4m/s with an uncertainty of +-5.3%(or 0.212)
confidence rating #$&*: 3
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
#$&*
"
Self-critique (if necessary):
------------------------------------------------
Self-critique rating:
@&
What is the maximum possible distance?
What is the minimum possible time interval?
What do you get when you calculate velocity based on these quantities?
By what percent does this result differ from the 4 m/s velocity you calculated?
*@
#*&!#*&!
&#Your work looks good. See my notes. Let me know if you have any questions. &#